Question

If the roots of the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are equal, then $a^2+b^2+c^2=$

• A. a + b + c
• B. 2a + b + c
• C. 3abc
• D. ab + bc + ca

Answer 1

Answer: (D)

Given equation is
$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$
$\Rightarrow \left(x^2-bx-ax+ab\right)+\left(x^2-cx-bx+bc\right)+\left(x^2-ax-cx+ac\right)=0$
$\Rightarrow \left[x^2-\left(a+b\right)x+ab\right]+\left[x^2-\left(b+c\right)x+bc\right]+\left[x^2-\left(a+c\right)x+ac\right]=0$
$\Rightarrow 8x^2-2\left(a+b+c\right)x+ab+bc+ca=0$
Since roots are equal
$\therefore$ We have
$b^2-4ac=0\Rightarrow b^2=4ac$
$4{\left(a+b+c\right)}^2=12\left(ab+bc+ca\right)$
${\left(a+b+c\right)}^2=3\left(ab+bc+ca\right)$
$a^2+b^2+c^2=ab+bc+ca.$